fskilnik wrote:
GMATH practice exercise (Quant Class 11)
If \(\,{{x - 1} \over {x - 3}}\,>\,{{x - 2} \over {x - 4}}\,\) is equivalent to \(\,a < x < b\,\), the value of \(b-a\) is:
(A) 0.25
(B) 0.5
(C) 0.75
(D) 1
(E) 1.25
Let´s see a "non-standard" (but absolutely rigorous) approach, very interesting for high-level candidates:
\({{x - 1} \over {x - 3}} > {{x - 2} \over {x - 4}}\,\,\,\left( * \right)\,\,\,\, \Leftrightarrow \,\,\,\,\,a < x < b\)
\(? = b - a\)
\(\left( * \right)\,\,\,\,\,\mathop \Leftrightarrow \limits^{ \cdot \left( {x - 3} \right)\left( {x - 4} \right)} \,\,\,\,\left\{ \matrix{\\
\,\left( {x - 1} \right)\left( {x - 4} \right) > \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,{\rm{if}}\,\,\,\,\left( {x - 3} \right)\left( {x - 4} \right) > 0\,\,\,\left( {\rm{I}} \right) \hfill \cr \\
\left( {x - 1} \right)\left( {x - 4} \right) < \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,{\rm{if}}\,\,\,\,\left( {x - 3} \right)\left( {x - 4} \right) < 0\,\,\,\left( {{\rm{II}}} \right) \hfill \cr} \right.\)
\(\left( {\rm{I}} \right)\,\,\,\,\left( {x - 1} \right)\left( {x - 4} \right) > \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\,{x^2} - 5x + 4 > {x^2} - 5x + 6\,\,\,{\rm{impossible!}}\)
\(\left( {{\rm{II}}} \right)\,\,\,\,\left( {x - 1} \right)\left( {x - 4} \right) < \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\,{x^2} - 5x + 4 < {x^2} - 5x + 6\,\,\,{\rm{always}}\,\,\left[ {\,{\rm{for}}\,\,\left( {x - 3} \right)\left( {x - 4} \right) < 0\,} \right]\,\,!\,{\rm{!}}\)
\(\left( {x - 3} \right)\left( {x - 4} \right) < 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,3 < x < 4\)
\(? = 4 - 3 = 1\)
The correct answer is (D).
We follow the notations and rationale taught in the
GMATH method.
Regards,
Fabio.